1. ## negative exponent elimination

Paper says that "negative exponents can be eliminated by replacing a negative power with its reciprocal" with the example of

(x^-3)(x^2)
------------
(x^-7)

becoming

(x^2)(x^7)
-----------
(x^3)

reducing out to x^6. how would the be done with (x^5)(X^-8) (no given denominator), just one numerator and denominator
x^5
----
x^-4
and two numerators and denominators ( i know i'm not using those terms right, but i think you get what i mean)
(x^4)(x^-6)
------------
(x^-5)(x^-2)

thanks once again

2. [quote=ScottPaulsen1;68439]Paper says that "negative exponents can be eliminated by replacing a negative power with its reciprocal" with the example of

(x^-3)(x^2)
------------
(x^-7)

becoming

(x^2)(x^7)
-----------
(x^3)

reducing out to x^6. how would the be done with (x^5)(X^-8) (no given denominator), just one numerator and denominator
x^5
----
x^-4
and two numerators and denominators ( i know i'm not using those terms right, but i think you get what i mean)
(x^4)(x^-6)
------------
(x^-5)(x^-2)

Ok, man you need to go to this site it has everything you need

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3. Originally Posted by ScottPaulsen1
Paper says that "negative exponents can be eliminated by replacing a negative power with its reciprocal" with the example of

(x^-3)(x^2)
------------
(x^-7)

becoming

(x^2)(x^7)
-----------
(x^3)

reducing out to x^6. how would the be done with (x^5)(X^-8) (no given denominator), just one numerator and denominator
x^5
----
x^-4
and two numerators and denominators ( i know i'm not using those terms right, but i think you get what i mean)
(x^4)(x^-6)
------------
(x^-5)(x^-2)

thanks once again
Remember we can always take a number x and put a 1 under it: $\displaystyle x = \frac{x}{1}$

So
$\displaystyle x^5x^{-8} = x^5 \cdot \frac{x^{-8}}{1} = \frac{x^5}{x^8}$

In the case of $\displaystyle \frac{x^5}{x^{-4}} = \frac{x^5x^4}{1} = x^5x^4$

Finally:
$\displaystyle \frac{x^4x^{-6}}{x^{-5}x^{-2}} = \frac{x^4x^5x^2}{x^6}$

-Dan